# Practice Paper 1 Question 12

Let \(n < 10\) be a non-negative integer. How many integers from \(0\) to \(999\) inclusive have the sum of their digits equal to \(n\)? Give your answer in terms of \(n\). *Hint:* Try first for integers from \(0\) to \(99\).

## Related topics

## Warm-up Questions

How many \(3\) digit numbers, whose digits consist solely of even numbers, exist?

A ternary number consists of only \(2\)s, \(1\)s and \(0\)s. How many values can be represented by a \(7\) digit ternary number?

## Hints

- Hint 1You can use inspection (there are only 10 cases) for the case of max 2 digits to get an expression in terms of \(n\).
- Hint 2For the case of max 3 digits, what if you fix one digit to some value \(p\le n\)?
- Hint 3What must be the sum of the remaining max 2 digits in terms of \(p\) and \(n\)?
- Hint 4How many max 2 digits numbers have their digit sum equal to \(n-p\)?
- Hint 5How should we incorporate that for all values of \(p\)?

## Solution

For max \(99\) (max 2 digits) case one can observe, by inspection, that there are \(n+1\) numbers whose digit sum is \(n\).

When there are max 3 digits, let's fix one of the digits to \(p\le n\). The sum of the remaining max 2 digits must thus equal \(n-p,\) and we know there are \(n-p+1\) numbers with that property. Taking all possible values of \(p\) we get \[\begin{align}\sum_{p=0}^n (n-p+1)&=(n+1)\sum_{p=0}^n 1-\sum_{p=0}^n p\\&=\frac{(n+1)(n+2)}{2}.\end{align}\]

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